p chapter 7 solutions 3rd ed fall 2007

# p chapter 7 solutions 3rd ed fall 2007 - Probability, Third...

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Sheet1 Page 1 Probability, Third Edition By David J Carr & Michael A Gauger Published by BPP Professional Education Solution 7.1 (i) Pr (X>3.1 )=1 -F (3.1) =1 -0.9990 =0.0010 (ii) Pr (X<-1.4 ( 1.4) =1 -F (1.4) =1 -0.9192 )=F- (iii) Pr 0.4 (<X<2.2 )=F (2.2) -F (0.4) =0.9861 -0.6554 =0.3307 (iv) Pr (-1.7 <X<- 0.2 ( 0.2) ( 1.7) =(1 -F (0.2) - 1 -F (1.7) ) )=F- -F- )( (0.2) =0.9554 -0.5793 =0.3761 Solution 7.2 The moment generating function of Xis: .s t. X .. .. Hence the cumulant generating function of X is: Solutions to practice questions h Chapter 7 Mt() =exp .ht+.

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Sheet1 Page 2 s t XX (i) The expected value of Xis: []= s EX R ' X (ii) The variance of Xis: () X2 var XR = '(0) =s ' Solution 7.3 The moment generating function is: .. 2 16t2 (t - t )=exp -12t +. exp 8( 1.5) . .. .. Rt() =ln (Mt () )=ht+ (0) =h+ Probability Solutions to practice questions h Chapter 7
Sheet1 Page 3 Comparing this to the MGF of a normal distribution: .st . X .. .. We can see that XN (- . ~ 12,16) Solution 7.4 .X -15 . By Theorem 7.1, . .~ N(0,1) . Hence: . X -15 18 -15 ) )=-. . 0.3 1 0.6179 =0.3821 (i) Pr (X >18 =1 -Pr (X =18 1Pr .= .=1 -F( ) Err:520 . 10 10 . 315 X -15 27 -15 Mt() =exp .ht +.

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Sheet1 Page 4 (ii) Pr 3 ( X 27 =Pr . < < .. << ) . - F- . 1010 10 . (1 F(1.2) )=2 F(1.2) -1 =2 0.8849 -1 =0.7698 Err:520 415 X -15 4 -15 . .-- (iii) Pr (4 X 4)=Pr . < < .=F- ( 1.1) -( 1.9) -<< F- . 10 10 10 . (1 -F(1.1) )( 1 F(1.9) =F(1.9) -F(1.1) =0.9713 -0.8643 =0.1070 Err:520 --) Solution 7.5 The probability that a sample value ( X ) is less than 9 is: .X -10 9 -10 . F- Pr (X <9)=Pr .< Err:520
Sheet1 Page 5 . . This last value is found using linear interpolation: 0.7071 -0.7 F(0.7) +[(0.8) -F(0.7) ] F(0.7071) F 0.8 -0.7 Hence, the required probability is: (62C )(0.2399 )( 0.7601 )=0.2882 0 320,000 320,000 ... Solution 7.6 Let X1 and X2 be the sizes of the two claims, and let YX= 1 -X2 be the difference between these random variables. Then by Theorem 7.4 (with a1 =1 and a2 =-1 ), we have: YN~ ((1,800 -1,800) , (400 +400 ))=N (0 , 320,000) The probability that the claims differ by more than \$500 is: Pr (Y <-500)+Pr (Y >500)=2 Pr(Y >500) by symmetry We have: . Y -0 500 -0 . Pr (Y >500)=1 -Pr (Y =500)=1 -Pr = . .. . 1 (0.8839) =1 -0.8114 h BPP Professional Education Solutions to practice questions h Chapter 7 Probability

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Sheet1 Page 6 Hence the required probability is: Pr (Y <-500)+Pr (Y >500)=2 Pr (Y >500)=2 0.1886 =0.3772 Solution 7.7 Let Xi be the lifetime of the i-th bulb. A succession of n bulbs will produce light for a random time equal to:
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## This note was uploaded on 09/08/2008 for the course STT 490 taught by Professor Dudziak during the Spring '08 term at Michigan State University.

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p chapter 7 solutions 3rd ed fall 2007 - Probability, Third...

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